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Math Help - primes and unique factorization

  1. #1
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    primes and unique factorization

    Prove or disprove each of the following statements:

    a. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 - c^2).

    b. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 + c^2).

    c. If p is prime and p divides a and p divides (a^2 + b^2), then p divides b.

    help plz..dont know where to start..
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by empressA View Post
    Prove or disprove each of the following statements:

    a. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 - c^2).

    b. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 + c^2).

    c. If p is prime and p divides a and p divides (a^2 + b^2), then p divides b.

    help plz..dont know where to start..
    What do you know about primes? Let's start with that.
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  3. #3
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    I know that p is a prime only if its divisors are positive neg. 1 and pos.neg. p..

    i also know that if p is a prime and q is a prime then p divides q

    and if p is a prime number that divides bc then it p divides b and p divides c..
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by empressA View Post
    I know that p is a prime only if its divisors are positive neg. 1 and pos.neg. p..

    i also know that if p is a prime and q is a prime then p divides q

    and if p is a prime number that divides bc then it p divides b and p divides c..
    That second statement is scarily wrong. 2,3.
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  5. #5
    Senior Member Dinkydoe's Avatar
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    This question should rather be posted in the number-theory section.

    (a) is false: let p= 5, a= 1, b= 2, c= 3, d = 1

    Have you tried a few examples?

    (b) Find a counter-example for (b).

    (c) Is correct. If a\equiv 0 mod p then a^2+b^2\equiv b^2 mod p. Thus (a^2+b^2)\equiv 0 mod p \Rightarrow b \equiv 0 mod p.
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  6. #6
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    thanks dinky doe..

    drexel 128..this is the notes my lecturer gave..lol..at least that second and third line
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  7. #7
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    Quote Originally Posted by Dinkydoe View Post

    (c) Is correct. If a\equiv 0 mod p then a^2+b^2\equiv b^2 mod p. Thus (a^2+b^2)\equiv 0 mod p \Rightarrow b \equiv 0 mod p.



    we haven't reached congruence modular arithmetic
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  8. #8
    Senior Member Dinkydoe's Avatar
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    Empress. I can hardly believe that's what your lecturer gave.

    The very definition of prime-numbers is that they can't be divided by numbers other then themself and 1.

    Thus if p,q are primenumbers and p|q \Rightarrow p=q. I think that's what your lecturer meant.
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  9. #9
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    yea..thats what i have..but i took it as p / q..because he didn't differentiate between the sign between the p and the q..but i c now
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  10. #10
    Senior Member Dinkydoe's Avatar
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    we haven't reached congruence modular arithmetic
    Then try to prove the following:

    If p|a, this means a = np for some n\geq 0. So a^2 = n^2p^2.
    And if p|(a^2+b^2)\Rightarrow a^2+b^2 = pm

    Thus if a^2+b^2 = n^2p^2+ b^2 = pm. What does this say about b^2?
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  11. #11
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    Thus if a^2+b^2 = n^2p^2+ b^2 = pm. What does this say about b^2?[/quote]

    that b is zero
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  12. #12
    Senior Member Dinkydoe's Avatar
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    No:

    Observe that from the above equation follows: b^2 = p(m-n^2p)\Rightarrow p|b^2

    Now use the definition of prime-numbers.
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  13. #13
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    that would imply that p divides a^2 + b^2
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  14. #14
    Senior Member Dinkydoe's Avatar
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    that would imply that p divides a^2 + b^2
    No, that's what we assumed in the first place. We also assumed that p divides a.

    In the last post I showed you that from these assumptions follows that p divides b^2.

    The definition of prime-numbers gives you that p divides b.
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  15. #15
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    ok..i understand where you're coming from now. so since p / b^2 it also divides b..
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